What gas is given off when metals react with acid?

A quarter tonne of sulfuric acid was reacted with half a tonne of iron to fill

https://en.wikipedia.org/wiki/Jacques_Charles#First_hydrogen_balloon

It depends on the metal, but when you work with Aluminum as an example, the reaction becomes:

`color(blue)("2Al"(s))` `color(blue)(+)` `color(blue)(6"HCl"(aq) -> 2"AlCl"_3(aq))` `color(blue)(+)` `color(blue)(3"H"_2(g))`

Another way to see this is the depiction of the reaction mechanism. At a General Chemistry class, you do not need to know this mechanism, but maybe it'll help to get a "behind-the-scenes" on what happens.

What you should notice is that indeed, two `"Al"` atoms begin with react with `"HCl"`. Since aluminum has three unpaired electrons, the `"Al"` atoms react by donating singular, radical electrons to cleave the `"H"-"Cl"` bond, and `"H"_2` gas forms.

After three steps, we form two equivalents of `"AlCl"_3(aq)` and a total of three equivalents of `"H"_2(g)`.

As an interesting aside, if we had two more equivalents of `"HCl"`, the reaction would proceed one more step and peak out to be:

`color(blue)(2"Al"(s))` `color(blue)(+)` `color(blue)(8"HCl"(aq) -> 2"AlCl"_4^(-)(aq))` `color(blue)(+)` `color(blue)(3"H"_2(g) + "2H"^(+)(aq))`

Since aluminum now has no more valence electrons to donate, it must utilize its empty `2p_z` orbital to accept two electrons from `"Cl"^(-)`, thus leaving behind one equivalent of `"H"^(+)` instead of forming `"H"_2(g)`.